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(i) Name the element of 3d transition series which shows maximum number of oxidation states. Why does it show so? (ii) Which transition metal of 3d series has positive E^(@)(M^(2+)//M) value and why? (iii) Out of Cr^(3+) and Mn^(3+), which is a stronger oxidising agent and why? (iv) Name a member of the lanthanoid series which is well known to exhibit +2 oxidation state. (v) Complete the following equation : MnO_(4)^(-)+8H^(+)+5e^(-)to |
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Answer» Solution :(i) Mn shows the maximum number of oxidation states, from +2 to +7. This is because it can lose or share maximum number of electrons `(3d^(5)4s^(2))`. (ii) Copper has the POSITIVE value for `E^(@)(M^(2+)//M)`. This is because the high energy to transform Cu (s) to `Cu^(2+)` is not balanced by its hydration energy. (iii) Out of `Cr^(3+)` and `Mn^(3+),Mn^(3+)` is a STRONGER OXIDISING agent. That is `Mn^(3+)` has a tendency to get reduced to `Mn^(2+)`. This is because `Mn^(3+)` has d-orbital configuration `3d^(4)` while `Mn^(2+)` has a more stable half-filled configuration `3d^(5)`. On the other HAND, `Cr^(3+)` has no tendency to act as oxidising agent. In other words, it has no tendency to change to `Cr^(2+).Cr^(3+)` has d-orbital configuration as `d^(3)`. These three electrons occupy the three `t_(2g)` orbitals (splitting of d-orbitals takes place). It is again half-filled stable configuration. Thus, `Cr^(3+)` has no tendency to act as oxidising agent. (iv) Sm and Eu EXHIBIT +2 oxidation state. (v) `MnO_(4)^(-)+8H^(+)+5e^(-)to Mn^(2+)+4H_(2)O` |
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